solução telecomunicações2

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Soluti on of Homework S Problem No: 5.1.1 wc =  108 , kf =  105, kp =  25 FM: the instantaneous frequency is W; = 108 + 10 5m (t) , (Wj  ) min = 108 — 10 5 = 9.99 X 107  ra d / s  (Wi)max  = 108 + 10 5 = 1.001 X 108 ra d /s 10-3 The below figure shows that cycle is split into four equal parts of length ------ = 2.5 X 10-4  second . 4 The instantaneous frequency increases linearly from (wt)min t o (wt)max  , st ays the re and then decreases linearly back to (wt)min and st ays there for the last quarter -cycle. m(t) WO 8000 _ _  _ _  _ -8000 PM: In this cas e = 10 8 + 25m (t) w ith (m (t))m ax = —8000, (w;  ) min = 10 8 2 X 10 5 = 9.98 X 107  ra d /s (Wi)max = 108 + 2 X 105 = 1.002 X 108 ra d/s The above fi gur e shows that the w aveform in this case st ays at (wt)max for one qua rter -cycle shif ts back to wcfor the last quarter -cycle. Problem No: 5.1.2 

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Solution of Homework S

Problem No: 5.1.1

wc =   108 , kf =   105, kp =   25

FM: the instantaneous frequency is W; = 108 + 10 5m (t) , (Wj )min  = 108 — 10 5 = 9.99 X 10 7 ra d/s 

(Wi)max  = 108 + 10 5 = 1.001 X 108 ra d/s

10-3The below figure shows that cycle is split into four equal parts of length ------= 2.5 X 10-4  second .

4

The instantaneous frequency increases linearly from (wt)min to (wt)max , stays the re and thendecreases linearly back to (wt)min and stays there fo r the last qua rter -cycle.

m(t) WO

8000 __ _  __   _ _ 

-8000

PM: In this case = 108 + 25m (t) w ith (m (t))max  = —8000, (w; )min  = 10 8 — 2 X 10 5 = 9.98 X

107 ra d/s

(Wi)max  = 108 + 2 X 105 = 1.002 X 108 ra d/s

The above figure shows tha t the w avefo rm in this case stays at (wt)max for one qua rter -cycle shifts

back to wcfo r the last quarter -cycle.

Problem No: 5.1.2 

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(a) wc= 2n X 106 , kf   = 2000^, kp  = n/2.

FM:

The instantaneous frequency is ft = 106 + 10 00 m (t), (f t )min  = 106 — 1000 = 999 kHz  . As shown

In the below figure the instantaneous frequency of the FM wave increases linearly from 999 to 1001 kHz

over 1 0- 3 , then switches back to 999 kHz and repeats.

PM:

Since m(t) has jum p d iscontinu ities , the direc t approaches will be used. When one cycle of the sawtooth

is centered on the origin, m (t) =2000t over th at cycle. Hence

2000t durante esse ciclo. por isso n 

V pm ( 0 = COs[2ff X 106 t  + ^ m(t)]

= cos[2rc X 10 6 t  + — 20 00 t]

= c os [2^ (106 1 + 50 0 ) t ]

 As shown in the above figure, at the disco ntin uity there is a jum p of 2kp  = n , otherwise the carrier

frequency is constant at (1 06 + 500 ) Hz.

\t VW -

m (b)This equivalent to another PM signal with

f c = 1000.5 KH z and periodic rectangula r message tha t sw itches from  1 to — 1 . it is necessary

to keep kp less than n. Since those periodic signal jumps at those points of discontinuity A=

2.Otherw ise a larger kp will give rise to phase ambiguity when kpA> 2n.

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Problem No: 5.1.4 

wc = 10000^ and that over   |t| < 1

<Pem  ( 0 = 10c os 13000rc t

(a) If this were a PM signal w ith kp  = 1000, we would have 

V pm  (0 = 10 cos 1300 0^ t   = 10 cos[wct  + kpm(t)]

= 10 cos [1000nt + 100 0m (t )]

Clearly, m (t)=3nt over this interval.

(b) For an FM signal w ith k f   = 1000;

<Ppm  ( 0 =  Aeos wct  + k  ̂ I m (a )da0

= 10 cos 10000n t + 1000I

Here f ^ m ( a ) d a = 3n t and m (t)  = 3n over the interval.

nt 

m(a)da0

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Solution of HW:9

Problem No: 5.2-3 

m(t)  = 2cos1000t +  9 cos2000nt so B = 1 KHz

(a) A=10 ,wc = 106,K f   = 1000n,kp = 1 .Th e re fore ,

=Aco s (wct + kpm (t)j 

= 10 cos((106t) + 2cos1000t + 9cos2000nt) 

Now f^a m{a)da   = 2 s in 10 00 t /10 00 + 9 s in 2000n t  / (2000 n:), so

^ p m ( 0 =A cos(w ct + kf I m(a)da' ■ ' - a

= 10 cos((106)t + 2nsin1000t +   4.5 sin 200 0n t)

(b) P M =m ( t ) = —2000s m1000 t — 18000nsin2000nt,mp  = 2000 + 18000n:

Kfm„  1000V / = - L J L =   --------+ 9 0 0 0   = 9.32 kHz

2 ^ n

Hence BPM  = 2 (A / + B ) = 2 (9 .32 + 1 ) = 20.64kHz

Kfmnmv  = 11 , V f = ̂ - ^ =   ( 500 ) (11 ) = 5.5kHz and BFM  = 2 (A / + B) = 2(5 .5 + 1) = 13kHz 

F 2n:

Problem No: 5.2-4 

w c = 2n x   106 ,q>EM(t )  = 10cos(wc t + 0 .1sm2000nt ) ,6 = 1 kH z

(b )9 (t) = wct + 0.1sin2000nt ,w i (t ) = w c + 200nsin 2000nt , Am  = 200^, A / = 100Hz

(c) A0 = 0.1

(d) B em   = 2 (A / + B) = 2(0.1 + 1) = 2.2Hz

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Problem No: 5.2-6

In this case, B = 3x1 0 3 = 3 kHz

PM:Kp  = 25 , mv  = = 8000 , V / = = (25)(8000)P ' P 10 J ’ > 2n 2n  

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Bpm  = 2(Af + B) =   2(31.82 + 3) = 69.66

FM:

 _  Kfm„  10sK f   = 10 5 ,m v  = 1 , V / = —— - =   ----- = 15.92kHz

1 ^ 2n 2n

Bfm   = 2 ( A / + B)   = 2(15.92 + 3) = 37.84 kH z

Problem No: 5.2-7 

m(t)  =  sin 2000nt , B = 1Khz,k f   = 200000^  ,and kp  = 10

(a) PM:

m ( t) = 200nsin2000nt ,mp  = 2000^

K„fmv  ( 10 ) (2000^ ) A f = pf v =   ± = 10kHz

2 ^ 2 ^

Bpm  = 2 (A / + B) = 2(10 + 1) = 22kHz

FM:

Kfm„  200000^mv = 1 , V f = —— - =   ------------- = 100kHz.

y 2n   2 ^

Thus Bfm   = 2 (A / + B ) = 2 (100 + 1) = 202kHz

(b) m(t) = 2sin2000nt ,B = 1Khz 

PM:

m(t) = 4000ncos2000nt ,mp  = 4 0 0 0 ^

K„fm„  ( 10 ) (4000^ ) A f = pr p =   ---------- = 20kHz

 J  2 ^ 2 ^

31.84 kH z

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Bpm  = 2 (A / + B) = 2(20 + 1) = 42kHz 

FM:

Kfm„  400000^mv = 2  , V f = —— - =   -------------- = 200kHz.

F 2^ 2^

Thus Bfm   = 2 (A / + B ) = 2 (200 + 1) = 402kHz

(c)

m ( t) =  sin 4000nt , B = 2k Hz 

PM:

m (t) = 4000n:cos4000n:t   , mp  = 4 0 0 0 ^

K„fm„  ( 10 ) (4000^ ) A f = pr p =   --------- - = 20kHz

 J  2 ^ 2 ^

Bpm  = 2 (A / + B) = 2(20 + 2) = 44kHz 

FM:

Kfm„  200000^mv = 2  , V f = - V ^ =   ------------- = 100kHz.

F 2 ^ 2 ^

Thus Bfm   = 2 (A / + B ) = 2 (100 + 2 ) = 204 kH z

(d)

Doubling the amplitude of m (t) roughly doubles the bandwidth of the both FM and PM. Doubling

the frequency of m(t) (i.e. expanding the spectrum o f M (^ ) by a facto r  2 )has hardly any effect on the

FM bandwidth. However, it roughly doubles the bandwidth of PM, indicating that the PM spectrum is

sensitive to the shape of the baseband spectrum. The FM spectrum is relatively insensitive to the nature

of the spectrum M (^ ).

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Solution of HW 10

Problem 5.3.1

 As shown in the graph, th e NBFM generators f  c1  = 2 0 0 KH z and = 9 to 1 0 H z .The final WBFM

should have f  c4  = 9 6   MHZ and Af 4  = 2 0 kHz .The total factor o f frequency mu ltiplication needed is

Mi. M2 =— =2000 to 2222.Because only frequency doubles are available, w e find tha t Mt . M2=2 1 1 And4fi

 A/1 = =9.7655 Hz. Now M1=2 n1  , M1 =2 n2, n1+n2=11, f c2=2 n1 f c1  and f c4=2 n1 f c 3. In o rde r to find

f Lo  , there are thre e possible relationships: f c3  = f  c2  + f Lo, f Lo_ f c2.Each should be tested to dete rm ine

the one tha t will req uire 9 X 1 06  << f Lo << 1 0 X 1 06  .First we tes t f  c3  = f  c2  —f Lo.This case

Leads to 96 MHz=fc4  = 2 n2fc 3  = 2 n2 (f c 2  —/ lo ) = 2 n 1+n2 fc 1  — 2 n2 fa   = 211  (2 0 0 X 1 03 ) — 2n2  f ^

Thus we havefLo=2 _n2 (4 . 09 6  X 1 08  — 9 .6  X 1 0 8) = 2 _ n2( 3 . 1 3 6  X 1 08).

In this case, i f n2=5 then f Lo = 9 .8   MHz, which is in the desired range. We won't test the other cases

since this one works. Thus, the fina l design M1=64, M2=32 and f Lo=9.8 M Hz. This give sf c2  = 2 n1 f c1  =

1 2 .8  MHz, A f  2  = M 1 . A/1 = 6  2 5 H z , f c3  = ( f c2  —f Lo) = 1 2 .8  —9 .8  = 3 MHz. A f  3  = 62 5 Hz .This

band pass filter used will be centered at 3 M Hz.

Problem 5.4.2

Given tha t f c = 1 0 kH z , A f = 1 kH z and the message is periodic square wave of period

resulting FM signal simply switches instantaneous frequ enc y from 11kHz to 9 kHz and back over

one pe riod. Thus

<PFM( t ) = A co s [2  0  0  07Tt + 2 0  0  0 nt ]

over any given half period. Now, after the ideal differe ntiato r—

<Pfm ( t ) =  — ( 2  0  0  0  0 n + 2 0  0  07t)A co s [2  0  0  0 n t + 2 0  0  0 n t]

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Next after the envelope detector, the ou tput w ill be a periodic square wave p roportional to (2 0 0 0 0n +

2 0 0 0n )A , with a dc offset. Afte r dc blocking, the is a periodic square wave prop ortion al to m(t).This is

illustrated in below figure:

Problem 5.4.3

s ( t) = 2 c o s [ 1 0 7 nt + 2 s in (2 0 0 0n t  + 0 . 3 n ) — 3 nco s( 1 0 0 t) ]

(a) The baseband bandw idth o f this angle -m od ula ted signal is B= 2 000n = 1 ZcHz.2n

 Also ( ) ( ) ( )

The refore Aw = 43 0 0n and A / = 2 . 1 5 ZcHz . Thus BFM = 2 (A / + B ) = 2 ( 2 . 1 5 + 1 ) =

6.3 kH z

(b) Since s(t) is an angle -m od ula ted signal w ith a constant am plitude of 2 , the o utp ut o f an ideal

envelope detecto r wou ld be just a constant.

(c) If s(t) is first differen tiated ,

S(t ) = (2 )(—1) [107n + 40 00n cos(2000nt + 0.3n) + 300n s in(10 0t) ] * s in[107 n t

+ 2 s in(2000nt + 0.3n) —3ncos(100t) ]

Thus the o utp ut of the ideal envelope dete ctor would be

2 [ 1 0 7n + 40 00n co s( 2 0 0 0n t + 0 . 3 n ) + 3 0 0n s in ( 1 0 0 t) ]

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. . 40007TSin(2 00 07Tt + 0. 37r) + 3 0 07Tsin(10 0t) I . 1  I- ■m(t)=------------ -------------------------------- (------1 = 2 0 c o s (2 0 0 0n t + 0 . 3 n ) + 1 . 5 s in ( 1 0 0 t)

kf   

(d) Clearly ,it is necessary to first differentiate the signal to obtain the message .If /e^=200n, then